Problem: Simplify the following expression: $x = \dfrac{-8q^2 + 64q - 96}{q - 2} $
Solution: First factor the polynomial in the numerator. We notice that all the terms in the numerator have a common factor of $-8$ , so we can rewrite the expression: $ x =\dfrac{-8(q^2 - 8q + 12)}{q - 2} $ Then we factor the remaining polynomial: $q^2 {-8}q + {12} $ ${-2} {-6} = {-8}$ ${-2} \times {-6} = {12}$ $ (q {-2}) (q {-6}) $ This gives us a factored expression: $\dfrac{-8(q {-2}) (q {-6})}{q - 2}$ We can divide the numerator and denominator by $(q + 2)$ on condition that $q \neq 2$ Therefore $x = -8(q - 6); q \neq 2$